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huh

callum can you show the steps please

(2, infinity)

x > 2

think you divided the wrong side there

oh yes sorry it's 2 lol, quite right.
\[(2,\infty)\]

rickjbr can you demonstrate how you got your answer please

you'll have imaginary numbers if the square root is allowed to go negative

thus not part of the real number system

callum was right in his calculation
4x - 8 > 0
4(x-2) > 0
x-2 > 0
x > 2

thanks for the help to both of your

np

how about this. 2x^2-3x+5...find and simplify the difference quotient

ah, right

(2(x+h)^2-3(x+h)+5-2x^2+3x-5)/h

(2(x^2+2xh+h^2)-3(x+h)+5-2x^2+3x-5)/h

(2x^2+4xh+2h^2-3x-3h+5-2x^2+3x-5)/h

(2h^2+4xh+3x)/h

is that the final answer

checking, a bit rusty

my options are a)4x+2h, b)4x+2h-3, c)4h+2x-3, d)4h+2x, e)4x+4h

i'll redo it, might've missed something

ok

ah, silly

that last step was it. instead of (2h^2+4xh+3x)/h
it should be (2h^2+4xh-3h)/h

simply factor out an h

[h(2h+4x-3)]/h
=2h+4x-3

that's it

probably better off posting on main forum, if i can help, i'll try

ok thanks for the help